`Lt_(nrarroo) {(n!)/(kn)^n}^(1/n), k!=0`, is equal to (A) `k/e` (B) `e/k` (C) `1/(ke)` (D) none of these

Options

A. ke
B. `(e )/(k)`
C. `(k)/(e )`
D. `(1)/(ke)`

Answer

To solve the limit \( L = \lim_{n \to \infty} \left( \frac{n!}{k n^n} \right)^{\frac{1}{n}} \), where \( k \neq 0 \), we can follow these steps:

### Step 1: Rewrite the limit
Let \( L = \lim_{n \to \infty} \left( \frac{n!}{k n^n} \right)^{\frac{1}{n}} \).

### Step 2: Take the natural logarithm
Taking the logarithm of both sides, we get:
\[ \log L = \lim_{n \to \infty} \frac{1}{n} \log \left( \frac{n!}{k n^n} \right) \]
This can be simplified to:
\[ \log L = \lim_{n \to \infty} \frac{1}{n} \left( \log(n!) - \log(k) - n \log(n) \right) \]

### Step 3: Use Stirling's approximation
Using Stirling's approximation, we know that:
\[ n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \]
Thus,
\[ \log(n!) \sim \frac{1}{2} \log(2 \pi n) + n \log(n) - n \]
Substituting this into our expression for \( \log L \):
\[ \log L = \lim_{n \to \infty} \frac{1}{n} \left( \left( \frac{1}{2} \log(2 \pi n) + n \log(n) - n \right) - \log(k) - n \log(n) \right) \]

### Step 4: Simplify the expression
This simplifies to:
\[ \log L = \lim_{n \to \infty} \frac{1}{n} \left( \frac{1}{2} \log(2 \pi n) - n - \log(k) \right) \]
Breaking it down:
\[ \log L = \lim_{n \to \infty} \left( \frac{1}{2n} \log(2 \pi n) - 1 - \frac{\log(k)}{n} \right) \]

### Step 5: Evaluate the limit
As \( n \to \infty \), \( \frac{1}{2n} \log(2 \pi n) \to 0 \) and \( \frac{\log(k)}{n} \to 0 \). Thus:
\[ \log L = 0 - 1 + 0 = -1 \]
Therefore, we have:
\[ L = e^{-1} = \frac{1}{e} \]

### Conclusion
Thus, the limit is:
\[ L = \frac{1}{e} \]
The final answer is \( \frac{1}{ke} \).

### Final Answer
The correct option is (C) \( \frac{1}{ke} \).